Is it valid to use operations on both sides before inequality is proven? As part of a bigger proof, I am trying to prove the inequality:
$$\frac{ab}{a^2 + b^2}< \frac{1}{2}$$
Is the following proving method correct?
$$\frac{ab}{(a^2)+(b^2)} <\frac{1}{2}$$
$$2ab < (a^2 + b^2)$$
$$0 <(a-b)^2$$
$a \neq b$ thus $a - b \neq 0$ and every number squared is not negative thus $(a-b)^2$ is positive thus bigger than zero
My concern with this proof is that I am not sure if I am allowed to do operations on both sides before inequality is proven. 
 A: What you've shown is $$\frac{ab}{a^2 + b^2} < \frac{1}{2} \implies 0 < (a-b)^2$$
But what you are (apparently) required to show is $$0 < (a-b)^2 \implies \frac{ab}{a^2 + b^2} < \frac{1}{2}$$
So the correct proof is basically the reverse of what you've written, namely:
Suppose $a ≠ b$. Then 
$$\begin{align} 0 < (a-b)^2  &\implies  2ab < (a^2 + b^2)\\
& \implies\frac{ab}{(a^2)+(b^2)} <\frac{1}{2}
\end{align}$$
$\square$
A: It is fundamentally correct, but not  well phrased. Actually, using the basic rule:

Let $A,B,C,D$ be real numbers and suppose $B,D$ are positive, then
$\dfrac AB<\dfrac CD\;$ if and only if $\;AD<BC$.

In your case the hypotheses on the denominators are satisfied. We deduce instantly the chain of equivalences
$$\frac{ab}{a^2 + b^2}< \frac{1}{2}\iff2ab <a^2+b^2\iff  a^2+b^2-2ab=(a-b)^2> 0$$
which is true, except if $a=b$.
A: Your proof is not mathematically incorrect, but it's written in a non-standard way that many people would consider incorrect. The conventional way to prove an equality or inequality is to start with a known sentence and work your way, step by step, to the sentence you want to prove. Therefore, you have two options:


*

*reverse the steps of your proof so that it goes the opposite way, or

*insert the word "because" between each step and the next, to show that your proof goes the opposite way from what is conventional.


(By the way, if I were you, I would also include more intermediate steps.)
If you choose the first option, your proof will look something like this:

Suppose we have two numbers $a > 0$ and $b > 0$, with $a \ne b$. We will prove that $\frac{ab}{a^2 + b^2}< \frac{1}{2}$. The proof is as follows.
$$
\begin{align*}
0 &< (a - b)^2 \\
0 &< a^2 - 2 a b + b^2 \\
2 a b &< a^2 + b^2 \\
\frac{2 a b}{a^2 + b^2} &< 1 \\
\frac{a b}{a^2 + b^2} &< \frac12
\end{align*}$$

If you choose the second option, your proof will look something like this:

Suppose we have two numbers $a > 0$ and $b > 0$, with $a \ne b$. Then $\frac{ab}{a^2 + b^2}< \frac{1}{2}$, because
$$
\begin{align*}
\frac{2 a b}{a^2 + b^2} &< 1, &\text{because} \\
2 a b &< a^2 + b^2, &\text{because} \\
0 &< a^2 - 2 a b + b^2, &\text{because} \\
0 &< (a - b)^2.
\end{align*}
$$

I definitely recommend the first option over the second option. The second option looks awkward because of all the "because"s, and yet we can't remove the "because"s because the proof would be more difficult to understand if we did. It's much easier to just write it in the conventional direction in the first place.
A: Other answers correct the form of this proof by demonstrating that the steps of the proof also imply each other in the reverse order. One could also correct this proof by converting it into a proof by contradiction and maintaining the order of steps.
As noted in the comments, the proposition is not true when $a = b$. Below I show that $ \frac{ab}{a^2+b^2} \leq 1 / 2$.
$$\frac{ab}{a^2+b^2} > 1 / 2 \implies 2ab > a^2 + b^2 \implies 0 > (a-b)^2$$
As you noted, $x^2 \geq 0$ for all real $x$, so the original assumption cannot be true.
I feel that this presentation improves the readability of the proof because it is read in the same order that it was most likely discovered.
A: As Nelver's answer highlights, you've technically proven the converse of what we want (i.e. the opposite implication). That said, this is something that is done a lot of the time by many people and in some sense, it's a more natural approach, but you have to be careful about how you write it up. One thing you can do is to say that all of the steps you've taken are reversible.
The other way to write this, which is more common, is to use equivalences (if and only ifs). So you could write \begin{align*} 
\frac{ab}{a^2+b^2}<\frac{1}{2} \iff 2ab<a^2+b^2 \iff 0<(a-b)^2
\end{align*} which is true since $a\neq b$. The $\iff$ is saying that the statement holds if and only if the statement on the right holds.
A: Yes your method is correct now you have to write it correctly. As we have $a^{2} + b^{2} -2ab \ge 0$ we can deduce $a^{2} + b^{2} \ge 2ab$ as if $a,b \ne 0$ as $a^{2} + b^{2} >0 $ we have $1 \ge 2 \frac{ab}{a^{2} + b^{2}}$ which proves the inequalities.
